This area combines coordinate geometry (lines, circles, parametric curves), sequences and series (arithmetic, geometric, sigma notation), proof by induction, and vectors in 2D and 3D.
Straight line: y − y₁ = m(x − x₁). Gradient m = (y₂−y₁)/(x₂−x₁). Perpendicular gradients: m₁m₂ = −1. Midpoint, distance formula. Circle: (x−a)² + (y−b)² = r², centre (a,b), radius r. Tangent perpendicular to radius at point of contact. Find intersection of line and circle by substitution. Parametric equations: x = f(t), y = g(t). Convert to Cartesian by eliminating t. dy/dx = (dy/dt)/(dx/dt). Area under parametric curve: ∫y(dx/dt)dt.
Arithmetic series: Sₙ = n/2(2a+(n−1)d) = n/2(a+l). Geometric series: Sₙ = a(1−rⁿ)/(1−r). Sum to infinity (|r|<1): S∞ = a/(1−r). Sigma notation: Σ from r=1 to n. Standard results: Σr = n(n+1)/2, Σr² = n(n+1)(2n+1)/6. Proof by induction (P4): base case → inductive step → conclusion. Vectors: position vectors, direction, magnitude. Scalar product a·b = |a||b|cosθ = a₁b₁+a₂b₂+a₃b₃. Perpendicular vectors: a·b = 0. Vector equation of line: r = a + td. Finding intersection point of two lines.
Proof by induction has three steps: (1) Base case: verify the statement is true for n=1 (or the starting value). Show both sides equal the same thing. (2) Inductive step: assume the statement is true for n=k (this is the inductive hypothesis). Then show it must also be true for n=k+1. This usually involves substituting n=k+1 into one side, then using the assumption for n=k to manipulate it until it matches. (3) Conclusion: state that since it\'s true for n=1, and if true for n=k it\'s true for n=k+1, then by the principle of mathematical induction it\'s true for all positive integers n. The logic is like dominoes — the base case knocks over the first, and the inductive step ensures each one knocks over the next.
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